You have found the following ages (in years) of all 5 tigers at your local zoo: $ 23,\enspace 10,\enspace 16,\enspace 6,\enspace 16$ What is the average age of the tigers at your zoo? What is the standard deviation? You may round your answers to the nearest tenth.
Answer: Because we have data for all 5 tigers at the zoo, we are able to calculate the population mean $({\mu})$ and population standard deviation $({\sigma})$ To find the population mean , add up the values of all $5$ ages and divide by $5$ $ {\mu} = \dfrac{\sum\limits_{i=1}^{{N}} x_i}{{N}} = \dfrac{\sum\limits_{i=1}^{{5}} x_i}{{5}} $ $ {\mu} = \dfrac{23 + 10 + 16 + 6 + 16}{{5}} = {14.2\text{ years old}} $ Find the squared deviations from the mean for each tiger. Age $x_i$ Distance from the mean $(x_i - {\mu})$ $(x_i - {\mu})^2$ $23$ years $8.8$ years $77.44$ years $^2$ $10$ years $-4.2$ years $17.64$ years $^2$ $16$ years $1.8$ years $3.24$ years $^2$ $6$ years $-8.2$ years $67.24$ years $^2$ $16$ years $1.8$ years $3.24$ years $^2$ Because we used the population mean $({\mu})$ to compute the squared deviations from the mean , we can find the variance $({\sigma^2})$ , without introducing any bias, by simply averaging the squared deviations from the mean $ {\sigma^2} = \dfrac{\sum\limits_{i=1}^{{N}} (x_i - {\mu})^2}{{N}} $ $ {\sigma^2} = \dfrac{{77.44} + {17.64} + {3.24} + {67.24} + {3.24}} {{5}} $ $ {\sigma^2} = \dfrac{{168.8}}{{5}} = {33.76\text{ years}^2} $ As you might guess from the notation, the population standard deviation $({\sigma})$ is found by taking the square root of the population variance $({\sigma^2})$ ${\sigma} = \sqrt{{\sigma^2}}$ $ {\sigma} = \sqrt{{33.76\text{ years}^2}} = {5.8\text{ years}} $ The average tiger at the zoo is 14.2 years old. There is a standard deviation of 5.8 years.